"Since Bram is back to a paid job the money will now (after March 2006) be used to help children in Uganda. This is the charity recommended by Vim's author. The money is used for a children centre in the south of Uganda, where AIDS has caused many victims. But at the same time donations increase Bram's motivation to keep working on Vim! "
Some terminal emulators are pretty weird about sending the Alt key. If you open vim, in insert mode, and type <C-v><Alt-s>, it'll show you what is keys are actually being sent. If it's not what you expect, try remapping it to something other than <Alt-s>.
Many comments suggest to use password managers like lastpass, 1pass, etc. But I think that may not be a good idea:
a. What if lastpass/1pass is compromised?
b. You have to login to retrieve your password, which is inconvenient.
I think the best solution to this is to make sure your passwords ONLY exist in your head, nowhere else. And to NOT reuse your passwords, you have to create a unique and reasonably strong one for each service.
So how do I remember all these unique and strong passwords? I create an algorithm which takes two parameters as inputs: my username and the domain of the service, it will do some simple manipulation of the inputs and give me a reasonably strong password. Hence, all you need to do is to remember your algorithm and use it to compute your password when you need it. Of course, you want the algorithm simple enough to be done in your head.
What the author mentioned is longest common SUBSEQUENCE (not substring). And it's true that LCS requires O(n^2) if only need to find ONE LCS, using dynamic programming. But if it's required to find ALL longest common subsequence, it definitely requires higher order. http://en.wikipedia.org/wiki/Longest_common_subsequence_prob...
I agree that the author should post the code he used to benchmark different languages. Otherwise, it's not convincing
Can you find the proof that finding all LCS is O(n^3)? The author was talking about "the standard algorithm for computing LCS". I was assuming the standard algorithm finds one LCS only.
Indeed, finding all LCSs requires exponential time: The strings "abcdefghijkl..." and "badcfehgjilk..." have the 2^(N/2) LCSs "[ab][cd][ef][gh][ij][kl]..." and no algorithm can ever run faster than the amount of time it takes to print its output.
